As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone.

Figure 1. The nondegenerate conic sections

Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines.

Figure 2. Degenerate conic sections

Identifying Nondegenerate Conics in General Form

In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below.

where [latex]A,B[/latex], and [latex]C[/latex] are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation.

You may notice that the general form equation has an [latex]xy[/latex] term that we have not seen in any of the standard form equations. As we will discuss later, the [latex]xy[/latex] term rotates the conic whenever [latex]\text{ }B\text{ }[/latex] is not equal to zero.

Finding a New Representation of the Given Equation after Rotating through a Given Angle

Until now, we have looked at equations of conic sections without an [latex]xy[/latex] term, which aligns the graphs with the x– and y-axes. When we add an [latex]xy[/latex] term, we are rotating the conic about the origin. If the x– and y-axes are rotated through an angle, say [latex]\theta[/latex], then every point on the plane may be thought of as having two representations: [latex]\left(x,y\right)[/latex] on the Cartesian plane with the original x-axis and y-axis, and [latex]\begin{align}\left({x}^{\prime },{y}^{\prime }\right)\end{align}[/latex] on the new plane defined by the new, rotated axes, called the x’-axis and y’-axis.

Figure 3. The graph of the rotated ellipse [latex]{x}^{2}+{y}^{2}-xy - 15=0[/latex]

We will find the relationships between [latex]x[/latex] and [latex]y[/latex] on the Cartesian plane with [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] on the new rotated plane.

Figure 4. The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle [latex]\theta [/latex].

The original coordinate x– and y-axes have unit vectors [latex]i[/latex] and [latex]j[/latex]. The rotated coordinate axes have unit vectors [latex]\begin{align}{i}^{\prime }\end{align}[/latex] and [latex]\begin{align}{j}^{\prime }\end{align}[/latex]. The angle [latex]\theta[/latex] is known as the angle of rotation. We may write the new unit vectors in terms of the original ones.

Figure 5. Relationship between the old and new coordinate planes.

Consider a vector [latex]u[/latex] in the new coordinate plane. It may be represented in terms of its coordinate axes.

Because [latex]\begin{align}u={x}^{\prime }{i}^{\prime }+{y}^{\prime }{j}^{\prime }\end{align}[/latex], we have representations of [latex]x[/latex] and [latex]y[/latex] in terms of the new coordinate system.

Example 2: Finding a New Representation of an Equation after Rotating through a Given Angle

Find a new representation of the equation [latex]2{x}^{2}-xy+2{y}^{2}-30=0[/latex] after rotating through an angle of [latex]\theta =45^\circ[/latex].

Show Solution

Find [latex]x[/latex] and [latex]y[/latex], where [latex]\begin{align}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \end{align}[/latex] and [latex]\begin{align} y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align}[/latex].

Because [latex]\theta =45^\circ[/latex],

[latex]\begin{align} &x={x}^{\prime }\cos \left(45^\circ \right)-{y}^{\prime }\sin \left(45^\circ \right) \\ &x={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right) \\ &x=\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}} \end{align}[/latex]

and

[latex]\begin{align}&y={x}^{\prime }\sin \left(45^\circ \right)+{y}^{\prime }\cos \left(45^\circ \right) \\ &y={x}^{\prime }\left(\frac{1}{\sqrt{2}}\right)+{y}^{\prime }\left(\frac{1}{\sqrt{2}}\right) \\ &y=\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}} \end{align}[/latex]

Substitute [latex]\begin{align}x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \end{align}[/latex] and [latex]\begin{align}y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \end{align}[/latex] into [latex]2{x}^{2}-xy+2{y}^{2}-30=0[/latex].

[latex]\begin{align} 2{\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)}^{2}-\left(\frac{{x}^{\prime }-{y}^{\prime }}{\sqrt{2}}\right)\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)+2{\left(\frac{{x}^{\prime }+{y}^{\prime }}{\sqrt{2}}\right)}^{2}-30=0\end{align}[/latex]

Simplify.

[latex]\begin{align}&2\frac{\left({x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }-{y}^{\prime }\right)}{2}-\frac{\left({x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }+{y}^{\prime }\right)}{2}+2\frac{\left({x}^{\prime }+{y}^{\prime }\right)\left({x}^{\prime }+{y}^{\prime }\right)}{2}-30=0 && \text{FOIL method} \\ &{x}^{\prime }{}^{2}{-2{x}^{\prime }y}^{\prime }+{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}+{x}^{\prime }{}^{2}+2{x}^{\prime }{y}^{\prime }+{y}^{\prime }{}^{2}-30=0 && \text{Combine like terms}. \\ &2{x}^{\prime }{}^{2}+2{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}=30 && \text{Combine like terms}. \\ &2\left(2{x}^{\prime }{}^{2}+2{y}^{\prime }{}^{2}-\frac{\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)}{2}\right)=2\left(30\right) && \text{Multiply both sides by 2}. \\ &4{x}^{\prime }{}^{2}+4{y}^{\prime }{}^{2}-\left({x}^{\prime }{}^{2}-{y}^{\prime }{}^{2}\right)=60 && \text{Simplify}. \\ &4{x}^{\prime }{}^{2}+4{y}^{\prime }{}^{2}-{x}^{\prime }{}^{2}+{y}^{\prime }{}^{2}=60 && \text{Distribute}. \\ &\frac{3{x}^{\prime }{}^{2}}{60}+\frac{5{y}^{\prime }{}^{2}}{60}=\frac{60}{60} && \text{Set equal to 1}. \end{align}[/latex]

Write the equations with [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] in the standard form.

[latex]\begin{align}\frac{{{x}^{\prime }}^{2}}{20}+\frac{{{y}^{\prime }}^{2}}{12}=1\end{align}[/latex]

This equation is an ellipse. Figure 6 shows the graph.

Figure 6

Writing Equations of Rotated Conics in Standard Form

Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex] into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the [latex]{x}^{\prime }[/latex] and [latex]{y}^{\prime }[/latex] coordinate system without the [latex]{x}^{\prime }{y}^{\prime }[/latex] term, by rotating the axes by a measure of [latex]\theta[/latex] that satisfies

We have learned already that any conic may be represented by the second degree equation

where [latex]A,B[/latex], and [latex]C[/latex] are not all zero. However, if [latex]B\ne 0[/latex], then we have an [latex]xy[/latex] term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle [latex]\theta[/latex] where [latex]\cot \left(2\theta \right)=\frac{A-C}{B}[/latex].

Example 3: Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term

Rewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[/latex] in the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system without an [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term.

Show Solution

First, we find [latex]\cot \left(2\theta \right)[/latex].

[latex]\begin{align}8{x}^{2}-12xy+17{y}^{2}=20\Rightarrow A=8,B=-12\text{ and }C=17\end{align}[/latex]

[latex]\begin{align} &\cot \left(2\theta \right)=\frac{A-C}{B}=\frac{8 - 17}{-12} \\ &\cot \left(2\theta \right)=\frac{-9}{-12}=\frac{3}{4} \end{align}[/latex]

Figure 7

[latex]\cot \left(2\theta \right)=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}[/latex]

So the hypotenuse is

[latex]\begin{gathered} {3}^{2}+{4}^{2}={h}^{2}\\ 9+16={h}^{2}\\ 25={h}^{2}\\ h=5\end{gathered}[/latex]

Next, we find [latex]\sin \text{ }\theta[/latex] and [latex]\cos \text{ }\theta[/latex].

[latex]\begin{align} \sin \theta &=\sqrt{\frac{1-\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1-\frac{3}{5}}{2}} \\ &=\sqrt{\frac{\frac{5}{5}-\frac{3}{5}}{2}} \\ &=\sqrt{\frac{5 - 3}{5}\cdot \frac{1}{2}} \\ &=\sqrt{\frac{2}{10}} \\ &=\sqrt{\frac{1}{5}} \\ &=\frac{1}{\sqrt{5}} \\[2mm] \cos \theta &=\sqrt{\frac{1+\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1+\frac{3}{5}}{2}} \\ &=\sqrt{\frac{\frac{5}{5}+\frac{3}{5}}{2}} \\ &=\sqrt{\frac{5+3}{5}\cdot \frac{1}{2}} \\ &=\sqrt{\frac{8}{10}} \\ &=\sqrt{\frac{4}{5}} \\ &=\frac{2}{\sqrt{5}} \end{align}[/latex]

Substitute the values of [latex]\sin \text{ }\theta[/latex] and [latex]\cos \text{ }\theta[/latex] into [latex]\begin{align}x={x}^{\prime }\cos \text{ }\theta -{y}^{\prime }\sin \text{ }\theta \end{align}[/latex] and [latex]\begin{align}y={x}^{\prime }\sin \text{ }\theta +{y}^{\prime }\cos \text{ }\theta \end{align}[/latex].

[latex]\begin{align}&x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ &x={x}^{\prime }\left(\frac{2}{\sqrt{5}}\right)-{y}^{\prime }\left(\frac{1}{\sqrt{5}}\right) \\ &x=\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}} \end{align}[/latex]

and

[latex]\begin{align} &y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \\ &y={x}^{\prime }\left(\frac{1}{\sqrt{5}}\right)+{y}^{\prime }\left(\frac{2}{\sqrt{5}}\right) \\ &y=\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}} \end{align}[/latex]

Substitute the expressions for [latex]x[/latex] and [latex]y[/latex] into in the given equation, and then simplify.

[latex]\begin{gathered}8{\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)}^{2}-12\left(\frac{2{x}^{\prime }-{y}^{\prime }}{\sqrt{5}}\right)\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)+17{\left(\frac{{x}^{\prime }+2{y}^{\prime }}{\sqrt{5}}\right)}^{2}=20 \\ 8\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left(2{x}^{\prime }-{y}^{\prime }\right)}{5}\right)-12\left(\frac{\left(2{x}^{\prime }-{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)+17\left(\frac{\left({x}^{\prime }+2{y}^{\prime }\right)\left({x}^{\prime }+2{y}^{\prime }\right)}{5}\right)=20 \\ 8\left(4{x}^{\prime }{}^{2}-4{x}^{\prime }{y}^{\prime }+{y}^{\prime }{}^{2}\right)-12\left(2{x}^{\prime }{}^{2}+3{x}^{\prime }{y}^{\prime }-2{y}^{\prime }{}^{2}\right)+17\left({x}^{\prime }{}^{2}+4{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}\right)=100 \\ 32{x}^{\prime }{}^{2}-32{x}^{\prime }{y}^{\prime }+8{y}^{\prime }{}^{2}-24{x}^{\prime }{}^{2}-36{x}^{\prime }{y}^{\prime }+24{y}^{\prime }{}^{2}+17{x}^{\prime }{}^{2}+68{x}^{\prime }{y}^{\prime }+68{y}^{\prime }{}^{2}=100 \\ 25{x}^{\prime }{}^{2}+100{y}^{\prime }{}^{2}=100 \\ \frac{25}{100}{x}^{\prime }{}^{2}+\frac{100}{100}{y}^{\prime }{}^{2}=\frac{100}{100} \end{gathered}[/latex]

Write the equations with [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] in the standard form with respect to the new coordinate system.

[latex]\begin{align}\frac{{{x}^{\prime }}^{2}}{4}+\frac{{{y}^{\prime }}^{2}}{1}=1\end{align}[/latex]

Figure 8 shows the graph of the ellipse.

Figure 8

Example 4: Graphing an Equation That Has No x′y′ Terms

Graph the following equation relative to the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system:

[latex]{x}^{2}+12xy - 4{y}^{2}=30[/latex]

Show Solution

First, we find [latex]\cot \left(2\theta \right)[/latex].

[latex]{x}^{2}+12xy - 4{y}^{2}=20\Rightarrow A=1,B=12,\text{ and }C=-4[/latex]

[latex]\begin{align}&\cot \left(2\theta \right)=\frac{A-C}{B} \\ &\cot \left(2\theta \right)=\frac{1-\left(-4\right)}{12} \\ &\cot \left(2\theta \right)=\frac{5}{12} \end{align}[/latex]

Because [latex]\cot \left(2\theta \right)=\frac{5}{12}[/latex], we can draw a reference triangle as in Figure 9.

Figure 9

[latex]\cot \left(2\theta \right)=\frac{5}{12}=\frac{\text{adjacent}}{\text{opposite}}[/latex]

Thus, the hypotenuse is

[latex]\begin{gathered} {5}^{2}+{12}^{2}={h}^{2}\\ 25+144={h}^{2}\\ 169={h}^{2}\\ h=13\end{gathered}[/latex]

Next, we find [latex]\sin \theta[/latex] and [latex]\cos \theta[/latex]. We will use half-angle identities.

[latex]\begin{align} \sin \theta &=\sqrt{\frac{1-\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1-\frac{5}{13}}{2}} \\ &=\sqrt{\frac{\frac{13}{13}-\frac{5}{13}}{2}} \\&=\sqrt{\frac{8}{13}\cdot \frac{1}{2}} \\ &=\frac{2}{\sqrt{13}} \\ \cos \theta &=\sqrt{\frac{1+\cos \left(2\theta \right)}{2}} \\ &=\sqrt{\frac{1+\frac{5}{13}}{2}} \\ &=\sqrt{\frac{\frac{13}{13}+\frac{5}{13}}{2}} \\ &=\sqrt{\frac{18}{13}\cdot \frac{1}{2}} \\ &=\frac{3}{\sqrt{13}} \end{align}[/latex]

Now we find [latex]x[/latex] and [latex]y[/latex].

[latex]\begin{align} &x={x}^{\prime }\cos \theta -{y}^{\prime }\sin \theta \\ &x={x}^{\prime }\left(\frac{3}{\sqrt{13}}\right)-{y}^{\prime }\left(\frac{2}{\sqrt{13}}\right) \\ &x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}} \end{align}[/latex]

and

[latex]\begin{align} &y={x}^{\prime }\sin \theta +{y}^{\prime }\cos \theta \\ &y={x}^{\prime }\left(\frac{2}{\sqrt{13}}\right)+{y}^{\prime }\left(\frac{3}{\sqrt{13}}\right) \\ &y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}} \end{align}[/latex]

Now we substitute [latex]\begin{align}x=\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\end{align}[/latex] and [latex]\begin{align}y=\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\end{align}[/latex] into [latex]\begin{align}{x}^{2}+12xy - 4{y}^{2}=30\end{align}[/latex].

[latex]\begin{align} &{\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)}^{2}+12\left(\frac{3{x}^{\prime }-2{y}^{\prime }}{\sqrt{13}}\right)\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)-4{\left(\frac{2{x}^{\prime }+3{y}^{\prime }}{\sqrt{13}}\right)}^{2}=30 \\ &\left(\frac{1}{13}\right)\left[{\left(3{x}^{\prime }-2{y}^{\prime }\right)}^{2}+12\left(3{x}^{\prime }-2{y}^{\prime }\right)\left(2{x}^{\prime }+3{y}^{\prime }\right)-4{\left(2{x}^{\prime }+3{y}^{\prime }\right)}^{2}\right]=30 && \text{Factor}. \\ &\left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+12\left(6{x}^{\prime }{}^{2}+5{x}^{\prime }{y}^{\prime }-6{y}^{\prime }{}^{2}\right)-4\left(4{x}^{\prime }{}^{2}+12{x}^{\prime }{y}^{\prime }+9{y}^{\prime }{}^{2}\right)\right]=30 && \text{Multiply}. \\ &\left(\frac{1}{13}\right)\left[9{x}^{\prime }{}^{2}-12{x}^{\prime }{y}^{\prime }+4{y}^{\prime }{}^{2}+72{x}^{\prime }{}^{2}+60{x}^{\prime }{y}^{\prime }-72{y}^{\prime }{}^{2}-16{x}^{\prime }{}^{2}-48{x}^{\prime }{y}^{\prime }-36{y}^{\prime }{}^{2}\right]=30 && \text{Distribute}. \\ &\left(\frac{1}{13}\right)\left[65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}\right]=30 && \text{Combine like terms}. \\ &65{x}^{\prime }{}^{2}-104{y}^{\prime }{}^{2}=390 && \text{Multiply}. \\ &\frac{{x}^{\prime }{}^{2}}{6}-\frac{4{y}^{\prime }{}^{2}}{15}=1 && \text{Divide by 390}. \end{align}[/latex]

Figure 10 shows the graph of the hyperbola [latex]\begin{align}\frac{{{x}^{\prime }}^{2}}{6}-\frac{4{{y}^{\prime }}^{2}}{15}=1\end{align}[/latex].

Figure 10

Identifying Conics without Rotating Axes

Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is

If we apply the rotation formulas to this equation we get the form

It may be shown that [latex]\begin{align}{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }\end{align}[/latex]. The expression does not vary after rotation, so we call the expression invariant. The discriminant, [latex]{B}^{2}-4AC[/latex], is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.

Key Equations

Key Concepts

• Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola.
• A nondegenerate conic section has the general form [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex] where [latex]A,B[/latex] and [latex]C[/latex] are not all zero. The values of [latex]A,B[/latex], and [latex]C[/latex] determine the type of conic.
• Equations of conic sections with an [latex]xy[/latex] term have been rotated about the origin.
• The general form can be transformed into an equation in the [latex]\begin{align}{x}^{\prime }\end{align}[/latex] and [latex]\begin{align}{y}^{\prime }\end{align}[/latex] coordinate system without the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term.
• An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section.

Glossary

angle of rotation

an acute angle formed by a set of axes rotated from the Cartesian plane where, if [latex]\cot \left(2\theta \right)>0[/latex], then [latex]\theta[/latex] is between [latex]\left(0^\circ ,45^\circ \right)[/latex]; if [latex]\cot \left(2\theta \right)<0[/latex], then [latex]\theta[/latex] is between [latex]\left(45^\circ ,90^\circ \right)[/latex]; and if [latex]\cot \left(2\theta \right)=0[/latex], then [latex]\theta =45^\circ[/latex]

degenerate conic sections

any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines.

nondegenerate conic section

a shape formed by the intersection of a plane with a double right cone such that the plane does not pass through the apex; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas

Section 7.6 Homework Exercises

1. What effect does the [latex]xy[/latex] term have on the graph of a conic section?

2. If the equation of a conic section is written in the form [latex]A{x}^{2}+B{y}^{2}+Cx+Dy+E=0[/latex] and [latex]AB=0[/latex], what can we conclude?

3. If the equation of a conic section is written in the form [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex], and [latex]{B}^{2}-4AC>0[/latex], what can we conclude?

4. Given the equation [latex]a{x}^{2}+4x+3{y}^{2}-12=0[/latex], what can we conclude if [latex]a>0?[/latex]

5. For the equation [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex], the value of [latex]\theta[/latex] that satisfies [latex]\cot \left(2\theta \right)=\frac{A-C}{B}[/latex] gives us what information?

For the following exercises, determine which conic section is represented based on the given equation.

6. [latex]9{x}^{2}+4{y}^{2}+72x+36y - 500=0[/latex]

7. [latex]{x}^{2}-10x+4y - 10=0[/latex]

8. [latex]2{x}^{2}-2{y}^{2}+4x - 6y - 2=0[/latex]

9. [latex]4{x}^{2}-{y}^{2}+8x - 1=0[/latex]

10. [latex]4{y}^{2}-5x+9y+1=0[/latex]

11. [latex]2{x}^{2}+3{y}^{2}-8x - 12y+2=0[/latex]

12. [latex]4{x}^{2}+9xy+4{y}^{2}-36y - 125=0[/latex]

13. [latex]3{x}^{2}+6xy+3{y}^{2}-36y - 125=0[/latex]

14. [latex]-3{x}^{2}+3\sqrt{3}xy - 4{y}^{2}+9=0[/latex]

15. [latex]2{x}^{2}+4\sqrt{3}xy+6{y}^{2}-6x - 3=0[/latex]

16. [latex]-{x}^{2}+4\sqrt{2}xy+2{y}^{2}-2y+1=0[/latex]

17. [latex]8{x}^{2}+4\sqrt{2}xy+4{y}^{2}-10x+1=0[/latex]

For the following exercises, find a new representation of the given equation after rotating through the given angle.

18. [latex]3{x}^{2}+xy+3{y}^{2}-5=0,\theta =45^\circ[/latex]

19. [latex]4{x}^{2}-xy+4{y}^{2}-2=0,\theta =45^\circ[/latex]

20. [latex]2{x}^{2}+8xy - 1=0,\theta =30^\circ[/latex]

21. [latex]-2{x}^{2}+8xy+1=0,\theta =45^\circ[/latex]

22. [latex]4{x}^{2}+\sqrt{2}xy+4{y}^{2}+y+2=0,\theta =45^\circ[/latex]

For the following exercises, determine the angle [latex]\theta[/latex] that will eliminate the [latex]xy[/latex] term and write the corresponding equation without the [latex]xy[/latex] term.

23. [latex]{x}^{2}+3\sqrt{3}xy+4{y}^{2}+y - 2=0[/latex]

24. [latex]4{x}^{2}+2\sqrt{3}xy+6{y}^{2}+y - 2=0[/latex]

25. [latex]9{x}^{2}-3\sqrt{3}xy+6{y}^{2}+4y - 3=0[/latex]

26. [latex]-3{x}^{2}-\sqrt{3}xy - 2{y}^{2}-x=0[/latex]

27. [latex]16{x}^{2}+24xy+9{y}^{2}+6x - 6y+2=0[/latex]

28. [latex]{x}^{2}+4xy+4{y}^{2}+3x - 2=0[/latex]

29. [latex]{x}^{2}+4xy+{y}^{2}-2x+1=0[/latex]

30. [latex]4{x}^{2}-2\sqrt{3}xy+6{y}^{2}-1=0[/latex]

For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation.

31. [latex]y=-{x}^{2},\theta =-{45}^{\circ }[/latex]

32. [latex]x={y}^{2},\theta ={45}^{\circ }[/latex]

33. [latex]\frac{{x}^{2}}{4}+\frac{{y}^{2}}{1}=1,\theta ={45}^{\circ }[/latex]

34. [latex]\frac{{y}^{2}}{16}+\frac{{x}^{2}}{9}=1,\theta ={45}^{\circ }[/latex]

35. [latex]{y}^{2}-{x}^{2}=1,\theta ={45}^{\circ }[/latex]

36. [latex]y=\frac{{x}^{2}}{2},\theta ={30}^{\circ }[/latex]

37. [latex]x={\left(y - 1\right)}^{2},\theta ={30}^{\circ }[/latex]

38. [latex]\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1,\theta ={30}^{\circ }[/latex]

For the following exercises, graph the equation relative to the [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] system in which the equation has no [latex]\begin{align}{x}^{\prime }{y}^{\prime }\end{align}[/latex] term.

39. [latex]xy=9[/latex]

40. [latex]{x}^{2}+10xy+{y}^{2}-6=0[/latex]

41. [latex]{x}^{2}-10xy+{y}^{2}-24=0[/latex]

42. [latex]4{x}^{2}-3\sqrt{3}xy+{y}^{2}-22=0[/latex]

43. [latex]6{x}^{2}+2\sqrt{3}xy+4{y}^{2}-21=0[/latex]

44. [latex]11{x}^{2}+10\sqrt{3}xy+{y}^{2}-64=0[/latex]

45. [latex]21{x}^{2}+2\sqrt{3}xy+19{y}^{2}-18=0[/latex]

46. [latex]16{x}^{2}+24xy+9{y}^{2}-130x+90y=0[/latex]

47. [latex]16{x}^{2}+24xy+9{y}^{2}-60x+80y=0[/latex]

48. [latex]13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-16=0[/latex]

49. [latex]4{x}^{2}-4xy+{y}^{2}-8\sqrt{5}x - 16\sqrt{5}y=0[/latex]

For the following exercises, determine the angle of rotation in order to eliminate the [latex]xy[/latex] term. Then graph the new set of axes.

50. [latex]6{x}^{2}-5\sqrt{3}xy+{y}^{2}+10x - 12y=0[/latex]

51. [latex]6{x}^{2}-5xy+6{y}^{2}+20x-y=0[/latex]

52. [latex]6{x}^{2}-8\sqrt{3}xy+14{y}^{2}+10x - 3y=0[/latex]

53. [latex]4{x}^{2}+6\sqrt{3}xy+10{y}^{2}+20x - 40y=0[/latex]

54. [latex]8{x}^{2}+3xy+4{y}^{2}+2x - 4=0[/latex]

55. [latex]16{x}^{2}+24xy+9{y}^{2}+20x - 44y=0[/latex]

For the following exercises, determine the value of [latex]k[/latex] based on the given equation.

56. Given [latex]4{x}^{2}+kxy+16{y}^{2}+8x+24y - 48=0[/latex], find [latex]k[/latex] for the graph to be a parabola.

57. Given [latex]2{x}^{2}+kxy+12{y}^{2}+10x - 16y+28=0[/latex], find [latex]k[/latex] for the graph to be an ellipse.

58. Given [latex]3{x}^{2}+kxy+4{y}^{2}-6x+20y+128=0[/latex], find [latex]k[/latex] for the graph to be a hyperbola.

59. Given [latex]k{x}^{2}+8xy+8{y}^{2}-12x+16y+18=0[/latex], find [latex]k[/latex] for the graph to be a parabola.

60. Given [latex]6{x}^{2}+12xy+k{y}^{2}+16x+10y+4=0[/latex], find [latex]k[/latex] for the graph to be an ellipse.